Question

The first term of arithmetic progression is $1$ and the sum of the first nine terms equal to $369.$ The first and the ninth term of a geometric progression conicide with the first and the ninth term of the arithmetic progression. Find the seventh term of the geometric progression.

Answer 1

$a,a+d,a+2d,.........+a+(n-1)d\to$ A.P.

$a=1$

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$S_9=369=\frac{9}{2}[2+(9-1\left)d\right]$

$41=1+4d$

$d=10$

$a+(n-1)d$

$a_9=1+8\times 10=81$

$A,AR,A{R}^2..........,A{R}^{N-1}\to$G.P.

$A,............,A{R}^8$

$A=1\to A{R}^8=81\Rightarrow R^8=81$

$R^8=3^4\Rightarrow R\sqrt{3}$

$\therefore A{R}^6=\left(1\right)(\sqrt{3}{)}^6=1(3{)}^3=27$